Using the xor swap algorithm
void xorSwap (int* x, int* y) {
if (x != y) { //ensure that memory locations are different
*x ^= *y;
*y ^= *x;
*x ^= *y;
}
}
Why the test?
The test is to ensure that x and y have different memory locations (rather than different values). This is because (p xor p) = 0
and if both x and y share the same memory location, when one is set to 0, both are set to 0.
When both *x and *y are 0, all other xor operations on *x and *y will equal 0 (as they are the same), which means that the function will set both *x and *y set to 0.
If they have the same values but not the same memory location, everything works as expected
*x = 0011
*y = 0011
//Note, x and y do not share an address. x != y
*x = *x xor *y //*x = 0011 xor 0011
//So *x is 0000
*y = *x xor *y //*y = 0000 xor 0011
//So *y is 0011
*x = *x xor *y //*x = 0000 xor 0011
//So *x is 0011
Should this be used?
In general cases, no. The compiler will optimize away the temporary variable and given that swapping is a common procedure it should output the optimum machine code for your platform.
Take for example this quick test program written in C.
#include <stdlib.h>
#include <math.h>
#define USE_XOR
void xorSwap(int* x, int *y){
if ( x != y ){
*x ^= *y;
*y ^= *x;
*x ^= *y;
}
}
void tempSwap(int* x, int* y){
int t;
t = *y;
*y = *x;
*x = t;
}
int main(int argc, char* argv[]){
int x = 4;
int y = 5;
int z = pow(2,28);
while ( z-- ){
# ifdef USE_XOR
xorSwap(&x,&y);
# else
tempSwap(&x, &y);
# endif
}
return x + y;
}
Compiled using:
gcc -Os main.c -o swap
The xor version takes
real 0m2.068s
user 0m2.048s
sys 0m0.000s
Where as the version with the temporary variable takes:
real 0m0.543s
user 0m0.540s
sys 0m0.000s